[SOLVED] Why is my awk script producing extra spacing between the first and second column?

mystupidquestion · 10-09-2011, 12:33 PM

I wrote the following awk program:

Code:

cat "$fileToProcess" | awk '!(/LINUX|.inux/ || /\%/ || /.verage/ || /^$/)\
  {
    ("date --date="$1 $2" +%T") | getline correctTime;
      printf "%s\t", correctTime;
      for (i=3; i<NF; i++) printf "%s\t", $i;
      printf "\n"
      close("date --date="$1 $2" +%T")
    }'

to parse a file. The program works as expected, but I'm curious as to why the spacing between column 1 and column 2 is greater than the other columns. Here is a sample of the output:

Code:

20:10:01        all     19.31   0.00    2.90    0.24    0.00    
20:20:01        all     10.30   0.00    1.97    0.22    0.00    
20:30:01        all     11.36   0.00    2.26    0.19    0.00    
20:40:01        all     12.04   0.00    2.37    0.20    0.00

colucix · 10-09-2011, 12:42 PM

It's not an awk problem: it is the way the TAB separator works. By default TAB is 8-spaces length, the first string is 8-characters so that a complete TAB is printed out immediately after that. Look at the spacing of your output:

Code:

-------8-------8-------8-------8-------8-------8-------8-------8
20:10:01        all     19.31   0.00    2.90    0.24    0.00    
20:20:01        all     10.30   0.00    1.97    0.22    0.00    
20:30:01        all     11.36   0.00    2.26    0.19    0.00    
20:40:01        all     12.04   0.00    2.37    0.20    0.00
-------8-------8-------8-------8-------8-------8-------8-------8

Every field is followed by a variable number of spaces to fill the 8-characters TAB separator (included the last one).

mystupidquestion · 10-09-2011, 12:51 PM

ic. Thanks!